Question

${\lim }_{x\to 3}\frac{\sqrt{5x+1}-\sqrt{7x-5}}{\sqrt{7x+4}-\sqrt{5x+10}}=$

• A. 0
• B. $-\frac{5}{4}$
• C. $\frac{5}{4}$
• D. $\infty$

Answer 1

Answer: (B)

${\lim }_{x\to 3}\frac{\sqrt{5x+1}-\sqrt{7x-5}}{\sqrt{7x+4}-\sqrt{5x+10}}=\frac{0}{0}$
That is, indeterminate form.
Multiply both the numerator and the denominator with $(\sqrt{5x+1}+\sqrt{7x-5})(\sqrt{7x+4}+\sqrt{5x+10})$
$\Rightarrow {\lim }_{x\to 3}\frac{[5x+1-(7x-5)](\sqrt{7x+4}+\sqrt{5x+10})}{(7x+4-5x-10)(\sqrt{5x+1}+\sqrt{7x-5})}$
$\Rightarrow {\lim }_{x\to 3}\frac{(-2x+6)(\sqrt{7x+4}+\sqrt{5x+10})}{(2x-6)(\sqrt{5x+1}+\sqrt{7x-5})}$
$={\lim }_{x\to 3}\frac{-(\sqrt{7x+4}+\sqrt{5x+10})}{(\sqrt{5x+1}+\sqrt{7x-5})}$
$=\frac{-(\sqrt{25}+\sqrt{25})}{(\sqrt{16}+\sqrt{16})}=\frac{-10}{8}=\frac{-5}{4}.$